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It may help to think of both cohort and case-control studies being limited, just in opposite ways. A case-control study cannot give you prevalence (risk) of disease, but a cohort study cannot give you prevalence of exposure. Only a cross-sectional study could give you both.

This is because prevalence (risk) is calculated from probabilities, and those probabilities in your study need to match what's in the actual population in order for you study to give good estimates. The moment you fix these probabilities to something other than reality, that limits what can be calculated from your study. For example, we force the probability of disease to 50% when we do a 1:1 case-control study. We force the probability of exposure to 50% when we do a 1:1 cohort study. These rarely match the true proportion in the population.

Here's an example. The true prevalences per group are shown in the table below from a cross-sectional study. | |${Disease}^+$|${Disease}^-$|$Total$|$P({Disease}^+)$ | |--------------|------------:|------------:|------:|-------------------------:| |${Exposure}^+$| 500| 1500| 2000| 25%| |${Exposure}^-$| 750| 1500| 2250| 33%| |$Total$ | 1250| 3000| 4250| 29%| |$P({Exposure}^+)$| 40%| 50%| 47%| |

In a cohort study, you are fixing the number in each exposure group, say 1200 in each group. Thus, your study can't estimate the probability of being exposed anymore: you forced it to be 50%. Notice how the risk of disease is unchanged (far column), but the prevalences of exposure are all different now. | |${Disease}^+$|${Disease}^-$|$Total$|$P({Disease}^+)$ | |--------------|------------:|------------:|------:|-------------------------:| |${Exposure}^+$| 300| 900| 1200| 25%| |${Exposure}^-$| 400| 800| 1200| 33%| |$Total$ | 700| 1700| 2400| 29%| |$P({Exposure}^+)$| 43%| 53%| 50%| |

In a case-control study, you are fixing the number of cases and controls, say 1200 in each group. Thus, your study can't estimate the probability of disease (risk). Notice how the prevalence of exposure is the correct, but the risk of disease is not.

Why?

RR is based on risks, and risks are population probabilities. We often estimate these probabilities using counts from our studies, but those estimates need to be valid approximations.

By definition, $RR=\dfrac{Risk_{{Exposure}^+}}{Risk_{{Exposure}^-}},$

where $Risk_{{Exposure}^+}=P({Disease}^+|{Exposure}^+)$, the probability of diseased given that one is exposed,

and $Risk_{{Exposure}^-}=P({Disease}^+|{Exposure}^-)$, the probability of being exposed given one is diseased.

Here's a generic cohort study. We picked $b$ and $d$ (e.g. 1200 people each), and then we measure $a$ and $c$ in the study (e.g. 300 and 400 respectively). I've left the rest of the table blank, because it's unneeded.

Looking at the table, you can estimate $Risk_{{Exposure}^+}=P({Disease}^+|{Exposure}^+)\approx \frac{a}{b}$, because $b$ is a constant, we measured $a$ in relation to $b$, and we do not need anything from the rest of the table. If I wind up picking a different $b$, then $\frac{a}{b}$ will stay roughly the same, because $a$ depends on $b$. You could view Row 1 as its own little experiment, independent of Row 2. Row 2 could have $d$ = 1 person or 1 billion people; it wouldn't change $a$, $b$, or the risk estimate. Similarly for $Risk_{{Exposure}^-}\approx \frac{c}{d}$.

[Delving into the statistical weeds, when $b$ and $d$ are fixed (aka given), we have two independent binomial distributions, $Bin(n, p)$. The expected value of a binomial distribution is $E[Bin(n, p)]=np$. For the exposed group, $n=b$ and $p=Risk_{{Exposure}^+}$. Since $a$ is our estimate of the expected value, we have $a\approx E[Bin(b,Risk_{{Exposure}^+})]=b\cdot Risk_{{Exposure}^+}$. Therefore, $Risk_{{Exposure}^+}\approx \frac{a}{b}$. Similarly for the non-exposed group.]

Now think about a case-control table. Here, we're given $y$ and $z$ as sample sizes for diseased and not diseased people, and from them we measure the number exposed, $w$ and $x$. I've added more detail to match the information we had for the cohort study.

The simplest "risk" to calculate here is the prevalence of exposure given disease status: $P({Exposure}^+|{Disease}^+)\approx \frac{w}{y}$. Compare to the cohort study, where the simplest risk was the risk of disease given exposure: $P({Disease}^+|{Exposure}^+)\approx \frac{a}{b}$. The previous answers have explained why $\frac{w}{y}\neq \frac{a}{b}$.

I suspect your question is actually: why can't we estimate $Risk_{{Exposure}^+}$ with $\frac{w}{w+x}$?

Unlike in the cohort design, we aren't given $w+x$; instead, we're given $y$ and $z$. Both $w$ and $x$ are measured quantities, and they are measured depending on $y$ and $z$. Now, we don't have a separate little experiment in the top row; instead, it's inextricably linked to the bottom Total row. We have to take that bottom row into account.

Remember that we picked $y$ and $z$, so we could pick any other numbers. Say we pick $z^*$, instead, where $z^*\neq z$, and from that we measure $x^*$ instead of $x$. Since $z^*\neq z$, then we would expect that $x^*\neq x$. Therefore, $w+x^*\neq w+x$ and so $\frac{w}{w+x^*}\neq \frac{w}{w+x}$. Thus we would get a different "risks" simply by choosing different sample sizes. The true risk, of course, does not change based on our selections. ||${Disease}^+$|${Disease}^-$|$Total$| |-|-|-|-| |${Exposure}^+$|$w$|$x^*$ ($\neq x$)|$w+x^*$| |${Exposure}^-$|$y-w$||$y+z^*-w-x^*$| |$Total$|$y$|$z^*$ ($\neq z$)|$y+z^*$|

[Back into the statistics, unlike $b$, $w+x$ is not constant. Therefore, the distribution in the top row is not $Bin(w+x,Risk_{{Exposure}^+})$. Since we cannot measure directly the risk, we'd have to use Bayes Theorem:$$Risk_{{Exposure}^+}=P({Disease}^+|{Exposure}^+)=\dfrac{P({Exposure}^+|{Disease}^+)P({Disease}^+)}{P({Exposure}^+)}.$$ We can estimate $P({Exposure}^+|{Disease}^+)\approx \frac{w}{y}$ and $P({Exposure}^+)\approx \frac{w+x}{y+z}$, because in both cases the binomial distribution holds. However, we cannot estimate $P({Disease}^+)\not\approx \frac{y}{y+z}$, because we chose $y$ and $z$, rather than measuring them from the experiment. Often we choose 1:1 groups, e.g. 1000 cases and 1000 controls. This would make $\frac{y}{y+z}=0.5$ even when the actual disease prevalence could be rare. If you know the disease prevalence from previous studies, then you can calculate risk and RR using it; but that information does not come from your case-control study (and isn't in the tables above).]